# How do you integrate int ( 1-3x)/((2x-1)(x+2)) using partial fractions?

$\textcolor{b l u e}{\int \frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} \mathrm{dx} = - \frac{1}{10} \cdot \ln \left(2 x - 1\right) - \frac{7}{5} \cdot \ln \left(x + 2\right) + C}$

#### Explanation:

From the given $\int \frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} \mathrm{dx}$
we use partial fraction with the variables $A$ and $B$
$\int \frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} \mathrm{dx} = \int \frac{A}{2 x - 1} \mathrm{dx} + \frac{B}{x + 2} \mathrm{dx}$

Let us determine the constants first
$\frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} = \frac{A}{2 x - 1} + \frac{B}{x + 2}$
$\frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} = \frac{A \left(x + 2\right) + B \left(2 x - 1\right)}{\left(2 x - 1\right) \left(x + 2\right)}$
$\frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} = \frac{A x + 2 A + 2 B x - B}{\left(2 x - 1\right) \left(x + 2\right)}$
Now we can set up the equations

$A + 2 B = - 3$
$2 A - B = 1$

Simultaneous solution results to:
$A = - \frac{1}{5}$
$B = - \frac{7}{5}$

Now, we can integrate
$\int \frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} \mathrm{dx} = \int \frac{A}{2 x - 1} \mathrm{dx} + \frac{B}{x + 2} \mathrm{dx}$
$\int \frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} \mathrm{dx} = \int \frac{- \frac{1}{5}}{2 x - 1} \mathrm{dx} + \frac{- \frac{7}{5}}{x + 2} \mathrm{dx}$
$\int \frac{1 - 3 x}{\left(2 x - 1\right) \left(x + 2\right)} \mathrm{dx} = - \frac{1}{10} \cdot \ln \left(2 x - 1\right) - \frac{7}{5} \cdot \ln \left(x + 2\right) + C$

God bless ..... I hope the explanation is useful.